Physics homework help

[Bbproz]

I am really stuck with my two physics problems remaining. My physics professors is horrible and so is the book we used, and after hours of attempting to solve these problems I cant. If anyone could help out... you'd be my hero <3.

1. A block of mass m1 = 4.50 kg on a frictionless inclined plane of angle 30.0° is connected by a cord over a massless, frictionless pulley to a second block of mass m2 = 2.60 kg hanging vertically.

What is the magnitude of the acceleration of each block?
What is the tension in the cord?

2. A "sun yacht" is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in everyday circumstances, it can be large enough tosend the spacecraft outward from the Sun on a cost-free but slow trip. Suppose that the spacecraft has a mass of 700 kg and receives a push of 10 N.

What is the magnitude of the resulting acceleration? .0142857143 m/s^2, got that part
If the craft starts from rest, how far will it travel in 1 day?
How fast will it then be moving?____ m/s

[TacTican]

I am really stuck with my two physics problems remaining. My physics professors is horrible and so is the book we used, and after hours of attempting to solve these problems I cant. If anyone could help out... you'd be my hero <3.

1. A block of mass m1 = 4.50 kg on a frictionless inclined plane of angle 30.0° is connected by a cord over a massless, frictionless pulley to a second block of mass m2 = 2.60 kg hanging vertically.

What is the magnitude of the acceleration of each block?
What is the tension in the cord?

2. A "sun yacht" is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in everyday circumstances, it can be large enough tosend the spacecraft outward from the Sun on a cost-free but slow trip. Suppose that the spacecraft has a mass of 700 kg and receives a push of 10 N.

What is the magnitude of the resulting acceleration? .0142857143 m/s^2, got that part
If the craft starts from rest, how far will it travel in 1 day?
How fast will it then be moving?____ m/s

In almost all physics classes, the TA will be one of your best friends. Schmooze with the TA. Bribe the TA. Get the TA to help you.

That being said, let's examine the problems. Usually, setting up the problem correctly is half the battle.

1. In force problems, it helps to draw free-body diagrams to visually orient yourself. Draw a picture of the pulley system with the weights. m2 is dangling off the side, therefore in our simple analysis it has two forces acting on it: the weight of the block and the tension of the connecting cord. The weight of the block pulls down, the tension of the cord pulls up, therefore they are pulling in opposite directions to each other. m1, resting on the inclined plane, is subject to three forces: the tension of the cord pulling it up the incline, the weight of the block pulling it directly down, and the normal force from Newton's third law preventing it from falling through the incline entirely.

However, since the mass m1 can only move in the plane of the incline (the weight of the block and the normal force will cancel each other, preventing the mass from moving off the surface of the incline), we need only analyze the direction of motion. The tension will pull the block up the incline. The weight will pull it down, but not all the weight. A little bit of trigonometry should tell you that W1*sin(30) is the amount of force pulling the block down the incline.

Putting these observations into mathematical form yields the following two equations:

Block 1: T - W1*sin(30) = m1*a
Block 2: T - W2 = m2*a

You have two equations and two variables (the tension T and the acceleration a). Here, I have used some simplifying assumptions: namely, that the tension of the cord will be the same between both blocks and the acceleration will also be the same between both blocks. To solve the problems, therefore, eliminate T and solve the equations for the acceleration a for the first part, and eliminate a and solve the equations for the tension T for the second part. The algebra I leave to you.

2. Okay, you got the first part. For the second part, you can treat the problem as a problem in kinematics: you have a constant acceleration supplied by a steady force. Therefore, you can use a kinematic equation to solve the problem. (Since I can never remember them off the top of my head I'll derive them instead, which is a faster and more powerful process for generating the equations needed to solve the problem.)

(d^2/dt^2)x(t) = a(t)

by definition: acceleration is the second derivative of position with respect to time. In this problem, however, the acceleration is independent of time, which makes everything easier. Integrate both sides once, and you get

(d/dt)x + v_0 = a*t

Here, the integration has removed an order of differentiation from the position side, left us with an integration constant in v_0, and given us a*t on the acceleration side. If this equation looks familiar, it should - it's just v + v_0 = a*t in another form. You're solving for v, you know that v_0 is zero because it was given to you, you solved for acceleration earlier, and you want to run for one day. Substitute, and the problem is now a matter of arithmetic.

Hope that helps.

[Lore Lord]

1) The force due to gravity is mg. The force due to gravity on the first block down the inclined plane is (4.5)(9.8)sin(30) = 22.05 N. This minus the tension will equal the total force on the block which is mass * acceleration. (4.5)a = 22.05 - T. For the second block there is no inclined plane so (2.6)a = (2.6)(9.8) - T. From the second equation we get T = 25.48 - 26a. Plugging that into the first equation we get (4.5)a = 22.05 - 25.48 + 26a. Solving this we get (21.5)a = 3.43. a = .1395 m/s^2. Solving the first equation for T, T = 21.42225 N.

2) X = X0 + V0t + (1/2)at^2. It starts from rest so X0 and V0 are 0. (1/2)(.0142857143)(86400(seconds in a day))^2 = 53321142.91 meters
V = V0 + at = 1234.2857 m/s

Hope this helps, in the future mmo champion prbly isn't the best place to ask for this sort of thing.

[Dacien]

Asking for help is fine, but in the future don't simply ask for answers. Closing this.